What is the solution for y^{''}+y^'-2y=3e^{-2t},y(0)=2,y^'(0)=1 ?

The solution for y^{''}+y^'-2y=3e^{-2t},y(0)=2,y^'(0)=1 is y=-e^{-2t}t+2e^t

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y^{''}+y^'-2y=3e^{-2t},y(0)=2,y^'(0)=1

y^{^{''}} + y^{^{'}} - 2 y = 3 e^{- 2 t}, y (0) = 2, y^{^{'}} (0) = 1

y = - e^{- 2 t} t + 2 e^{t}

Solution steps

y^{''} (t) + y^{'} (t) - 2 y = 3 e^{- 2 t}

Solve linear ODE:

y = - e^{- 2 t} t + 2 e^{t}

y = - e^{- 2 t} t + 2 e^{t}

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x, y (0) = 4, y^{^{'}} (0) = 8

y^{^{''}} - 25 y = e^{5 t}, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + 4 y = 4 t^{2}

(D^{2} - 4) y = 4 x - 3 e^{x}

y^{^{'''}} - y^{^{''}} - y^{^{'}} + 1 = 0

What is the solution for y^{''}+y^'-2y=3e^{-2t},y(0)=2,y^'(0)=1 ?
The solution for y^{''}+y^'-2y=3e^{-2t},y(0)=2,y^'(0)=1 is y=-e^{-2t}t+2e^t